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Problem Solving

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by bburton11 » Thu Jun 18, 2009 9:12 am
Two members of a club are to be selected to represent the club at a national meeting. If there are 190 different possible selections of the 2 members, how many members does the club have?

20
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by ghacker » Thu Jun 18, 2009 9:16 am
Answer is 20

First member can be selected in N ways and the second member can be selected in (N-1) ways , but you can arrange the two in 2! ways

190= N*(N-1)/2! ................. N(N-1) = 380 , N = 20
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by gmatplayer » Thu Jun 18, 2009 1:19 pm
n!/((n-k)!(k!)) = 190

n=?
k=2

n!/((n-2)!(2!)) = 190

n(n-1)=380

n=-19 or 20

n=20
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by [email protected] » Mon Jun 22, 2009 2:28 am
Thanks for the details- however if you could pls. explain how we got -

n(n-1) = 380.

Thanks.
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Problem Solving

by katz » Mon Jun 22, 2009 3:12 am
Here is how we arrive at n(n-1).

General Form
n! = n*(n-1)*(n-2)*(n-3)*(n-4)*(n-5)*...*1
I can shorten the above representation as follows:
n! = n*(n-1)*(n-2)! (by adding the "!" factorial notation)
C(n,r) = n!/[r!*(n-r)!]
C stands for combination

In our case, the problem is something like this

C(n,2) = n!/[2!*(n-2)!] ... I
Expanding the numerator in I and taking hint from the general form we get
= [n*(n-1)*(n-2)!] / [2! * (n-2)!]
Now we have (n-2)! in both the numerator and denominator which we can get rid off. Doing this we would have
= [n*(n-1)]/2! ... II
But from the problem II = 190 (R.H.S)
Therefore,
=> [n*(n-1)]/2! = 190
=> [n*(n-1)] = 380
=> n*(n-1) = 380
I am sure you are in control from here on.

Best of luck.
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